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Abstract

We consider a monotone increasing operator in an ordered Banach space having u− and u+ as a strong super- and subsolution, respectively. In contrast with the well-studied
case u+<u−, we suppose that u−<u+. Under the assumption that the order cone is normal and minihedral, we prove the
existence of a fixed point located in the order interval [u−,u+].

MSC:
47H05, 47H10, 46B40.

Keywords:

fixed point theorems in ordered Banach spaces

Research

It is an elementary consequence of the intermediate value theorem for continuous real-valued
functions f:[a1,a2]→R that if either

f(a1)>a1andf(a2)<a2(1)

or

f(a1)<a1andf(a2)>a2,(2)

then f has a fixed point in [a1,a2]. It is a natural question whether this result can be extended to the case of ordered
Banach spaces. A number of fixed point theorems with assumptions of type (1) are well
known; see, e.g., [[1], Section 2.1]. However, to the best of our knowledge, fixed point theorems with assumptions
of type (2) have not been known so far. In the present note, we prove the following
fixed point theorem of this type.

Theorem 1LetXbe a real Banach space with an order coneKsatisfying

(a) Khas a nonempty interior,

(b) Kis normal and minihedral.

Assume that there are two points inX, u−≪u+, and a monotone increasing compact continuous operatorT:[u−,u+]→X. Ifu−is a strong supersolution ofTandu+is a strong subsolution, that is,

Tu−≪u−andTu+≫u+,

thenThas a fixed pointu∗∈[u−,u+].

Here [u−,u+] denotes the order interval {u∈X:u−≤u≤u+}.

Theorem 1 generalizes an idea developed by the present authors in [2], where the existence of solutions to a certain nonlinear integral equation of Hammerstein
type has been shown.

Before we present the proof, we recall some notions. We write u≥v if u−v∈K, u>v if u≥v and u≠v, and u≫v if u−v∈K∘, where K∘ is the interior of the cone K.

A cone K is called minihedral if for any pair {x,y}, x,y∈X, bounded above in order there exists the least upper bound sup{x,y}, that is, an element z∈X such that

(1) x≤z and y≤z,

(2) x≤z′ and y≤z′ implies that z≤z′.

Obviously, a cone K is minihedral if and only if for any pair {x,y}, x,y∈X, bounded below in order there exists the greatest lower bound inf{x,y}. If a minihedral cone has a nonempty interior, then any pair x,y∈X is bounded above in order. Hence, sup{x,y} and inf{x,y} exist for all x,y∈X.

A cone K is called normal if there exists a constant N>0 such that x≤y, x,y∈K implies ∥x∥X≤N∥y∥X.

By the Kakutani-Krein brothers theorem [[3], Theorem 6.6] a real Banach space X with an order cone K satisfying assumptions (a) and (b) of Theorem 1 is isomorphic to the Banach space
C(Q) of continuous functions on a compact Hausdorff space Q. The image of K under this isomorphism is the cone of nonnegative continuous functions on Q.

An operator T acting in the Banach space X is called monotone increasing if u≤v implies Tu≤Tv.

Consider the operator Tˆ:[u−,u+]→X defined by

Tˆu:=sup{inf{Tu,u+},u−}.(3)

Since inf{Tu+,u+}=u+ and sup{u+,u−}=u+, u+ is a fixed point of the operator Tˆ. Similarly, one shows that u− is also a fixed point.

Proof For any v∈K, the maps u↦sup{u,v} and u↦inf{u,v} are continuous; see, e.g., Corollary 3.1.1 in [4]. Due to the continuity of T, it follows immediately that Tˆ is continuous as well. The operator Tˆ is monotone increasing since inf and sup are monotone increasing with respect to
each argument. Therefore, for any u∈[u−,u+], we have

u−=Tˆu−≤Tˆu≤Tˆu+=u+.

Let (un) be an arbitrary sequence in [u−,u+]. Since T is compact, (Tun) has a subsequence (Tunk) converging to some v∈X. From the continuity of Tˆ, it follows that the sequence (Tˆunk) converges to sup{inf{v,u+},u−}, thus, proving that the range of Tˆ is relatively compact. □

Lemma 3There existp±∈Xwith

u−≪p−≪p+≪u+

and

Tˆp−<p−,Tˆp+>p+.

Proof Due to Tu−≪u−, there is a δ>0 such that Bδ(u−−Tu−)⊂K∘. The preimage of Bδ(u−−Tu−) under the continuous mapping u↦u−Tu contains a ball Bϵ(u−). Hence, u−Tu≫0 holds for all u∈Bϵ(u−). By the same argument, u−Tu≪0 for all u∈Bϵ(u+). Choosing ϵ>0 sufficiently small, we can achieve that Bϵ(u−)∩Bϵ(u+)=∅.

Set p(t):={(1−t)u−+tu+|t∈[0,1]}. We choose t−∈(0,1) so small that p−:=p(t−)∈Bϵ(u−) and t+∈(0,1) so close to 1 that p+:=p(t+)∈Bϵ(u+). Then we have u−≪p−≪p+≪u+ and

Tp−≪p−,Tp+≫p+.

Due to p−≪u+ and Tp−≪p−, we have inf{Tp−,u+}=Tp−. Further, we obtain

sup{Tp−,u−}≤sup{p−,u−}=p−.

From Tp−≪p− it follows that there is an element z≪0 such that Tp−=p−+z. Assume that sup{Tp−,u−}=p−. Then we have sup{z,u−−p−}=0. However, in view of the Kakutani-Krein brothers theorem, u−−p−≪0 implies sup{z,u−−p−}≪0. Thus, it follows that sup{Tp−,u−}≠p− and, therefore, Tˆp−<p−. Similarly one shows that Tˆp+>p+. □

The main tool for the proof of Theorem 1 is Amann’s theorem on three fixed points
(see, e.g., [[5], Theorem 7.F and Corollary 7.40]):

Theorem 4LetXbe a real Banach space with an order cone having a nonempty interior. Assume there are four points inX,